Question

Let a vector \[ \vec a=\sqrt{2}\,\hat i-\hat j+\lambda \hat k,\ \lambda>0, \] make an obtuse angle with the vector \[ \vec b=-\lambda^2\hat i+4\sqrt{2}\hat j+4\sqrt{2}\hat k \] and an angle \( \theta \), \(\frac{\pi}{6}<\theta<\frac{\pi}{2}\), with the positive \(z\)-axis. If the set of all possible values of \( \lambda \) is \((\alpha,\beta)-\{\gamma\}\), then\(\alpha+\beta+\gamma\) is equal to ____________. 

Hint:

Always apply dot-product conditions before angular constraints to simplify vector problems.

Answer 1

Correct Answer: 5

Given vectors \(\vec{a}=\sqrt{2}\hat{i}-\hat{j}+\lambda\hat{k}\) and \(\vec{b}=-\lambda^2\hat{i}+4\sqrt{2}\hat{j}+4\sqrt{2}\hat{k}\), \(\lambda > 0\), we need to find conditions under which \(\vec{a}\) makes an obtuse angle with \(\vec{b}\).

The dot product \(\vec{a} \cdot \vec{b}\) must be less than zero for the angle to be obtuse: \(\vec{a} \cdot \vec{b} = \sqrt{2}(-\lambda^2) + (-1)(4\sqrt{2}) + \lambda(4\sqrt{2})\). Simplifying gives:

\[-\lambda^2\sqrt{2} - 4\sqrt{2} + 4\lambda\sqrt{2} < 0 \Rightarrow -\lambda^2\sqrt{2} + 4\lambda\sqrt{2} - 4\sqrt{2} < 0\]

Factoring out \(\sqrt{2}\):

\[-\lambda^2 + 4\lambda - 4 < 0\]

Solving the quadratic: \(\lambda^2 - 4\lambda + 4 = 0\) gives \(\lambda = 2\).

The parabola \(\lambda^2 - 4\lambda + 4\) is non-positive between roots; hence, \((0,2)\) for obtuse angle\)

Next, ensuring \(\frac{\pi}{6} < \theta < \frac{\pi}{2}\) with the positive \(z\)-ax\) using \(\cos\theta = \frac{\lambda}{|\vec{a}|}\):

\(|\vec{a}| = \sqrt{(\sqrt{2})^2 + (-1)^2 + \lambda^2} = \sqrt{2 + 1 + \lambda^2}\) gives:

\[\frac{\lambda}{\sqrt{3 + \lambda^2}} > \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\]

Cross-multiplying and squaring: 

\[\lambda^2 > \frac{3}{2}(3 + \lambda^2) \Rightarrow 2\lambda^2 > 9 + 3\lambda^2\]

Simplifies to \(-\lambda^2 > 9 \Rightarrow \lambda^2 < -9\), none exists but ensures consideration of \((2,\infty)\).

After observing both conditions, the solution set is \((0,2) \cup (2,\infty)\), avoiding a double root subse\)

\(\alpha = 0 \), \( \beta = 2 \), \( \gamma = 2\); hence, \(\alpha + \beta + \gamma = 4\).

Conclusion: Values of \(\lambda\) satisfying conditions provide \(0 + 2 + 2 = 4\).